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1- Introduction : La commande des machines électriques est l’une des applications des convertisseurs statiques. Cette commande nécessite l’association d’une machine (courant continu, synchrones,
asynchrones asynchrones ou autres) dont le fonctionnement est à une vitesse variable en lui conservant un couple optimum, à un convertisseur statique (redresseur, hacheur, gradateur, onduleur). En fait, le choix du moteur d’entraînement dépend du travail demandé, du lieu de travail et de la ce d’énergie dont on dispose, les contraintes sur les puissance à fournir. De même, la sour sour ce paramètres que l’on doit fournir et le prix de revient de l’ensemble déterminent le type du convertisseur à associer au moteur. Alors, on ambitionne d’étudier et d’analyser les possibilités d’association de convertisseur en vue de la commande.
Il est essentiel de connaître connaître les mises en en équation des machines à courant courant alternatif, continue et des convertisseurs convertisseurs pour en déduire les schémas schémas fonctionnels à partir desquels les commandes sont conçues. Les mises en équation des machines sont basée ba séess hab habit ituel uelle leme ment nt sur su r des de s tra t rans nsfo form rmati ations ons qui qu i exist exi stent ent sous so us diff di ffére érent ntes es varian var iante tess Enfin on peut mettre en en ouvre, vérifier toutes ces équations et avoir avoir une perspective de notre système système avant même de le réalisé grâce à un outille informatique très utiliser en électrotechnique qui y est matlab-Simulink, voici une brève description de ce dernier Simulink est un environnement environnement de schéma de principe principe pour la simulation multi domaine. Il prend en charge charge la conception de niveau système, système, simulation, génération génération automatique de de code, et un test en continu et la vérification des systèmes embarqués. embarqués. Simulink fournit un éditeur graphique, bibliothèques de blocs personnalisables personnalisables et solveurs pour la modélisation et la simulation des systèmes dynamiques. Il est intégré i ntégré avec MATLAB, vous permettant d'incorporer des algorithmes MATLAB dans des modèles et des résultats de simulation à l'exportation à MATLAB pour une analyse ultérieure. Comme objectifs du tp :
Nous essayerons essayerons de modélisé la machine à courant courant continue, machine machine asynchrone asynchrone en premier lieux, ensuite entamé les déférentes déférentes simulations concernent concernent ces machines machines ainsi que les convertisseurs associés. Ensuite nous allons analyser le comportement des caractéristiques (vitesse, courant ……..) des machines cités cités précédemment sans et avec couple résistant. Avant d’entamer la rédaction des équations un rappelle théorique sera bénéfique pour bien
assimiler les étapes qui seront étudiées ultérieurement.
Partie1 : convertisseur- machine à courant continu Principe de fonctionnement :
La vitesse de rotation d’un moteur à courant continu dépend de sa tension d’alimentation Le hacheur :
Un hacheur est un système électronique permettant de faire varier la vitesse d'un moteur à courant continu en faisant varier la tension moyenne d'alimentation du moteur Umot = Vcc 1
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Redresseur commandé :
Comme pour le hacheur, le redresseur permet de faire varier la tension moyenne du moteur. On fait varier la tension motrice en agissant sur l'angle d'amorçage des thyristors.
L'avantage du redresseur est qu'il transforme directement la tension alternative en tension continue variable ce qui représente un cout moins important par rapport au hacheur. Synoptique du system qui sera étudié à travers le tp :
Redresseur: permet de transformer une tension alternative en tension continue ondulée. Filtrage: Elimine les phénomènes d'ondulation de la tension en sortie du redresseur. Récupération: Système permettant de transformer l'énergie mécanique lors du freinage du
moteur en énergie calorifique dans le cas ou l'on utilise une résistance de dissipation comme système de freinage. Ces systèmes de récupération d'énergie assurent un freinage contrôlé du moteur. Hacheur: Permet de faire varier la tension moyenne du moteur donc la vitesse. 2-Les équations théoriques : a- Modélisation du moteur à courant continu :
Moteur à courant continu à excitation indépendante : Equation électrique :
= =
. . +
=
.
=
.
+
.
+
( -
. . )
=
(
-
.
-
Equation du couple électromagnétique :
=
*
Equation du mouvement du moteur entraine une charge :
)
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= ( -. - ) = ( . - - ) b- Mise en équation de la machine asynchrone :
b-1) Modèle de Park du MAS référentiel lié au champ tournant : Equations magnétiques :
Equations électriques :
Equation du couple électromagnétique :
Equation du mouvement :
Avant d’entamer la simulation nous allons devoir procéder à quelques étapes préliminaires dans le but est d’avoir une bonne organisation du travail qui sera effectuer durant le tp
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Pour la résolution de nos équations différentielles on a opté à la méthode de Runge-Kutta d’ordre 4, avec un pas de 1e-4.est la méthode qui sera utiliser pour la résolution r ésolution des
déférentes équations différentiels
3-Les schémas Matlab_Simulink permettant de résoudre les équations citées: 3-1- La machine à courant continu :
Avant d’introduire les schémas on doit définir les paramètres de : – –
La machine à courant continue Les déférents redresseurs statiques :
Paramètres de la MCC:
%paramétre d'une machine à courant continu à exitation exitat ion indépendante %fonction exitation indépendante Km=Mfd*if Vq=220; Rq=0.26; Kf=0.001; J=0.1; Km=0.6; Lq=0.0078; Cr=0; f=50; w=2*pi*f; Vmax=220*sqrt(2); T=1/f; alpha=50;
Paramètres du redresseur: %paramètres du redresseur f=50; w=2*pi*f; Vmax=220*sqrt(2);
Paramètres du hacheur: %paramètres du hacheur f=200; w=2*pi*f; Vmax=220*sqrt(2); T=1/f; alpha=50;
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Modèle (1): outils Simulink
1
2
iq 1
1 s
1/Lq
vq
-K-
Integrator
Gain
wr
Gain3 Rq
1 s
1/J
Gain2
Integrator1
2 cr
Gain1
Kf Gain4
-KGain5
Modèle (2): Matlab Fcn 1 iq 1 vq
u
MATLAB Function
1 s
MATLAB Fcn
Integrator
u
MATLAB Function
1 s
2
MATLAB Fcn1
Integrator1
wr
2 cr
MATLAB Fcn : (u(1)-Km*u(2)-Rq*u(3))/Lq MATLAB Fcn1 : (u(1)*Km-u(2)-Kf*u(3))/J
Modèle (3): Transfert Fcn 1 iq
1 vq
1
-K-
Lq.s+Rq Transfer Fcn
1 J.s+Kf
Gain
Transfer Fcn1
2 wr
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Simulation d’un pont redresseur triphasé à diodes: MATLAB Function MATLAB Fcn1
1 Alpha
MATLAB Function
max
MATLAB Fcn2
MinMax1 1
MATLAB Function
Product 2
MATLAB Fcn3
Vmax mi n MinMax
Schéma simplifié du pont redresseur à diodes :
t Clock
Gain
To Workspace Workspace
w
Alpha
Ured
ured To Works Workspace1
MATLAB Function
Vmax
Vmax
Constant Redresseur à diodes
MATLAB Fcn: 2*pi*floor(u(1)/(2*pi)) MATLAB Fcn1: sin (u(1)) MATLAB Fcn2: sin(u(1)-2*pi/3) MATLAB Fcn3: sin(u(1)-4*pi/3)
Ured
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Simulation d’un hacheur série : série :
Après simplification du system:
MATLAB Fcn: 2*pi*floor(u(1)/(2*pi))
3-1-1: Principaux résultants de simulation de la MCC: a ) Simulation de la MCC sans couple résistant : a-1) Tracage du courant courant d’induit et de la vitesse de rotation sans couple résistant :
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La courbe du courant en fonction du temps :
courant en fonction du temps 600
500
400
) A ( t 300 n a r u 200 o c 100
X: 0.8931 Y: 0.6104 0
-100 0
0. 1
0. 2
0. 3
0. 4
0. 5
0. 6
0. 7
0. 8
temps(S)
Courbe de la vitesse de rotation en fonction du temps :
vitesse en fonction du temps 400
350
) s 300 / d r ( n250 o i t a t o 200 r e d150 e s s e t i 100 v 50
X: 0.7212 Y: 366.4
0. 9
1
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courant durant l’intervalle de temps [0 ,0.4](s),et une fois que la machine attient son régime permanent sa vitesse devienne devienne constante, constante, l’équation du mouvement mouvement :Cem-Cr =J.
deviendra
Cem=f.wr
f.wr
Qui est u n couple faible et c’est pour ça qu’on voit que le courant I q a
diminué et reste presque nul.
Le courant avant qu’il atteigne son régime permanent il passe par un pique au régime
transitoire qui y est du au démarrage du moteur car pour faire tourner t ourner le rotor ce dernier doit consommer un courant très important pour vaincre son inertie a l’état statique ; jusqu’a
atteindre une vitesse constante et voir le courant se stabiliser au régime permanent On peut traduire cela par les équations suivantes : Cem-Cr =J.
f.wr
au régime permanent la vitesse sera constante constante donc :
Cem=f.wr
De même pour la vitesse elle passe par un régime transitoire (elle augmente) jusqu’a atteindre une valeur constante (régime permanent). La valeur maximale du courant au régime transitoire est : Iq = 567.1 A a-3) Explication du résultat:
= . . + . + = ( - . - - ) - - . = J = ( . . - - ) Au régime permanant =0et donc on aura : ( -. - 0 ( - . - 0
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a.4.La valeur maximal du courant au début du régime transitoire :
La valeur maximale du courant au régime transitoire est : Iq = 567.1 A a-4) explication du résultat :
( -. - ) (. - - )
Dans ce cas on préconise préconise la méthode de superposition : (Δiq(0)=0, Δwr(0)=0) par définition définition et en passant dans le domaine de laplace
Δvq(p)
*
k m
Δcr (p) (p)
R q+Lq*p
=
Δwr (p) (p)
-(J*p+k f f ) ) k m
Δiq(p)
Δvq(p)
0
k m -(J*p+k f f )
Δcr (p) (p)
D’ici on aura Δiq(p)=
) * Δvq(p) k m* Δcr (p)+ (p)+ (J*p+k f f)*
= K m²+ (R q+Lq*p)(J*p+k f f)
K m²+(R q+Lq*p)(J*p+k f f)
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3-1-2) Simulation de l’action d’un couple résistant qui entre entr e en action à un instant donné : (t=0.6s Cr=50N.m): a-1 : courbe du courant induit avec couple résistant: fig3: courant induit en fonction du temps 600
500
) 400 A ( t i u 300 d n i t n 200 a r u o c 100
X: 0.9296 Y: 84.36
0
-100 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
temps(S)
Interpretation : Le courant augumente consedirablement au moment de l’aplication du couple resistant , car c’est la charge qui éxige cette cett e augmentaion cela peut se traduit par la relation suivante : Ce =
Pe/(2.pi.n) = E'*I / 2.pi.n =K*I a-2 : courbe de la vitesse avec couple résistant à (t=0.6s) : fig4: vitesse en fonction du temps 400 350 300 250
) s / d200 r
X: 0.8985 Y: 330.1
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a-4 : Interprétation des resultats numériques :
= . . + . + = ( - . - - ) - - . = J = ( . . - - ) =0et donc on aura :
Au régime permanant
1 Lq
( m . iq - Cr - f r
if Cr f r
diq
dr
dt
dt
0 0
( q - m. r - q iq
1
0
0 0
(q - m . r - q iq ( m. iq - Cr - f r
Vq=220; Rq=0.26;
0.6wr +0.26i +0.26i q=220 ……. (1)
Kf=0.001; on aura donc Km=0.6;
0.6iq-50-0.001wr =0 ……… (2
Cr=50 ; De l’équation (2) on a wr =0.6/0.001 iq-50/0.001 =600 i q-50000
En remplaçant la valeur de w r dans l’équation (1) on obtient :
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3-2 : Simulation de l’ensemble MCC-redresseur-hacheur MCC-redresseur-hacheur :
Matlab Function: 2*pi*floor(u(1)/(2*pi))
a-1 : Traçage du courant induit sans couple résistant : fig5: courant en fonction du temps 600
500
400
) A
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a-2 : Traçage de la vitesse de rotation sans couple résistant : fig6: vitesse vite sse de rotati on en fonction fonction du temps temps 300
250
) s / d r 200 ( n o i t a t o 150 r e d e s 100 s e t i v
X: 1.84 Y: 254.6
50
0 0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
temps(S)
Interprétation :
Le rapport cyclique infecte la vitesse de rotation du moteur étudie en d’autres terme la vitesse
varie avec la variation du rapport cyclique alors, on peut conclure que le hacheur est très fiable pour la commande d’un moteur a courant continu.
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Interprétation :
Le courant augmente au moment de l’a pplication du couple couple résistant, pour pour
répondre aux
besoins de la machine. a-5 : traçage de la vitesse de rotation avec couple résistant à : (t=1.6 et Cr=50 N.m) : fig8: vitesse vite sse de rotati on en fonction fonction du temps temps 300
250
X: 1.879 Y: 218.9
) s / d r 200 ( n o i t a t o 150 r e d e s 100 s e t i v 50
0 0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
temps(S)
Interprétation :
La vitesse diminue au moment de l application du couple résistant et cela ce traduit par la relation suivante : Cem-Cr =f.wr Cem-Cr =f.wr du coup wr=(Cem-Cr)÷f
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Partie2 : convertisseur- machine asynchrone
Généralités à propos de la machine asynchrone :
La machine asynchrone, connue également sous le terme anglo-saxon de machine à induction, est une machine une machine électrique à courant alternatif sans connexion entre le stator le stator et le rotor. le rotor. Comme les autres machines électriques (machine à courant continu, machine continu, machine synchrone), synchrone) , la machine asynchrone est un convertisseur électromécaniqu convertisseur électromécaniquee basé sur l'électromagnétisme permettant 'électromagnétisme permettant la conversion conversion bidirectionnelle d'énergie d'énergie entre une installation électrique parcourue par un courant un courant électrique (ici alternatif) (ici alternatif) et et un dispositif mécanique. Cette machine est réversible et susceptible de se comporter, selon la source d'énergie, soit en « moteur » soit en « générateur », dans les quatre quadrants du plan couple-vitesse La machine asynchrone a longtemps été fortement concurrencée par la machine synchrone dans les domaines de forte puissance, jusqu'à l'avènement l'avènement de l 'électronique de puissance. Elle puissance. Elle est utilisée dans de nombreuses applications, notamment dans le transport le transport (métro, trains, propulsion des navires, automobiles électriques), dans l'industrie (machines-outils), (machines-outils), dans l'électroménager. l'électroménager. Elle Elle était à l'origine l 'origine uniquement utilisée en « moteur » mais, toujours grâce à l'électronique de puissance, elle est de plus en plus souvent utilisée en « génératrice » , par exemple dans les éoliennes.
Paramètres de la Machine asynchrone utilisé durant la simulation :
%alimentation fr=50; W=2*pi*fr;
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Vrefmax=10; m=21; Veo=220*(3/2)*sqrt(2);
Simulation d’un moteur asynchrone alimenté en tension : tension : Modèle de Park-Référentiel lié au champ tournant : Schéma détaillé du bloc MAS :
MATLAB Function
1
Vas
Va 2
Vds Vbs
MATLAB Function
1 s
MATLAB Fcn
Integrator
MATLAB Function
1 s
MATLAB Fcn1
Integrator1
idr
iar
MATLAB Fcn6 MATLAB Function
Mux
iqr
ibr
Terminator4
MATLAB Fcn7 Tetar
icr
Vb 3 Vc 4
MATLAB Function
1 s
MATLAB Fcn2
Integrator2
5
MATLAB Function
1 s
Cr
MATLAB Fcn3
Integrator3
Mux
Vcs Vqs Teta
Teta transformation de park
MATLAB Function 6 ws
MATLAB Fcn4 MATLAB Function
Terminator5 idsr
3
ia
ias iqsr
ib
Terminator2
1
Tetas
ic
Cem 1 s
4 iar
Terminator3 2
transformation inverse de park 1
transformation inverse de park1
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Bloc Transformation de Park :
1 Vas Va s
MATLAB Function
2
MATLAB Fcn8
Vbs Vb s
1 Vds
Mux Mu x
3 MATLAB Function
Vcs 4 Teta
Mu x
MATLAB Fcn9
2 Vqs
MATLAB Fcn : sqrt(2/3)*(u(1)*cos(u(4))+u(2)*c sqrt(2/3)*(u(1)*cos(u(4))+u(2)*cos(u(4)-2*pi/3)+u(3)*cos os(u(4)-2*pi/3)+u(3)*cos(u(4)-4*pi/3)) (u(4)-4*pi/3))
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Simulation de l’Onduleur triphasé commandé par MLI «Modulation «Modulat ion de Largeur d’Impulsion» : d’Impulsion» : Bloc onduleur à MLI :
1
vr
ca
To Works Workspace1
To Workspace
Vref Vref 1
Vref Vref 1
Vref Vref 2
Vref Vref 2
C1
C1
teta
Va
1 Va
teta
C2
2
Vref Vref max
Vref Vref 3
C2
Vref Vref 3
Vb
Vrefmax Tensions de référence référence
3
teta1
Vp
C3
C3
Onduleur à MLI 6
teta1
Veo
Vc
m
m
vp
vp To Works Workspace2
3 Vc
Veo 4
2 Vb
Onduleur
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Matlab Fcn : u(2)*sin(u(1)) Matlab Fcn1 : u(2)*sin(u(1)-2*pi/3) Matlab Fcn2 : u(2)*sin(u(1)-4*pi/3)
Bloc tension Porteuse :
1 teta1 Mu x
MATLAB Function MATLAB Fcn
2 m
3 Vpmax
Mux
MATLAB Function
1
MATLAB Fcn1 Fcn1
vp
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Matlab Fcn: (u(4)/3)*(2*u(1)-u(2)-u(3)) Matlab Fcn1: (u(4)/3)*(2*u(2)-u(1)-u(3)) Matlab Fcn2: (u(4)/3)*(2*u(3)-u(1)-u(2))
Bloc machine asynchrone assemblé à l’onduleur l’onduleur :
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a-1-1 : Tracé du courant statorique :
Figure du courant statorique en fonction du temps 80
60
40
20
) A
0
X: 0.9211 Y: 8.263
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a-1-3 : Tracé du couple électromagnétique : Fig 10 : couple électromagnétique en fonction du temps 300
250
200
) m . 150 N ( m e 100 c
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Calcul de la fréquence rotorique : f r=g*f s= ((1500-1498) /1500)*50=0 .07 Hz.
Avec, f s= 50 Hz
b : Simulation d’une alimentation directe de la MAS avec le couple résistant (à résistant (à t=0.5 et Cr=50N.m): Cr=50N.m): b-1 : Tracé du courant statorique et rotorique, du couple électromagnétique et de la vitesse de rotation en fonction du temps: b-1-1 : Tracé du courant statorique :
Fig 12 : courant statorique en fonction du temps
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b-1-3 : Tracé du couple électromagnétique :
Fi g 14 :co uple électro magnétique magnétique en fonction fonction du temps temps 300
250
200
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b-2 : Les valeurs du courant rotorique, statorique, la vitesse et le couple et le couple électromagnétique au régime permanent :
ias(A) iar(A) Cem(N.m) wr(rd/s)
Régime établi 13.08 9.715 51.46 146.5
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c-1-2 : Tracé du courant rotorique : fig16: courant rotorique en fonction du temps
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Interprétations :
Pour la fig15 on remarque une intense ondulation périodique.
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D-1-2 : Tracé du courant rototorique : fig20:
t ro tori
e en fonction du temps
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Interprétations :
De la figure 20 à l’instant d’insertion de couple résistant on remarqua l’apparition d’un
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